题目视图：

题目详解代码：

// 定义链表节点类
class ListNode {int val;ListNode next;ListNode(int x) {val = x;next = null;}
}public class PartitionList {public ListNode partition(ListNode pHead, int x) {// 创建两个虚拟头节点，分别用于存储小于 x 和大于等于 x 的节点ListNode smallDummy = new ListNode(0);ListNode largeDummy = new ListNode(0);// 分别指向两个新链表的当前节点ListNode smallTail = smallDummy;ListNode largeTail = largeDummy;// 遍历原链表ListNode current = pHead;while (current != null) {if (current.val < x) {// 如果当前节点的值小于 x，将其连接到 small 链表的尾部smallTail.next = current;smallTail = smallTail.next;} else {// 如果当前节点的值大于等于 x，将其连接到 large 链表的尾部largeTail.next = current;largeTail = largeTail.next;}// 移动到下一个节点current = current.next;}// 断开 large 链表的尾部，防止出现循环largeTail.next = null;// 将 small 链表和 large 链表连接起来smallTail.next = largeDummy.next;// 返回重新排列后的链表的头节点return smallDummy.next;}public static void main(String[] args) {// 创建测试链表 1 -> 4 -> 3 -> 2 -> 5 -> 2ListNode head = new ListNode(1);head.next = new ListNode(4);head.next.next = new ListNode(3);head.next.next.next = new ListNode(2);head.next.next.next.next = new ListNode(5);head.next.next.next.next.next = new ListNode(2);PartitionList solution = new PartitionList();int x = 3;// 调用 partition 方法进行重新排列ListNode newHead = solution.partition(head, x);// 打印重新排列后的链表ListNode current = newHead;while (current != null) {System.out.print(current.val + " ");current = current.next;}}
}

题目视图：

题目详解代码：

package Demo1_28;/*** Created with IntelliJ IDEA.* Description:* User：Lenovo* Date:2025-01-28* Time:20:04*/
// 定义链表节点类
class ListNode {int val;ListNode next;ListNode(int x) {val = x;next = null;}
}public class PalindromeLinkedList {public boolean isPalindrome(ListNode A) {if (A == null || A.next == null) {return true;}// 步骤 1：找到链表的中间节点ListNode slow = A;ListNode fast = A;while (fast.next != null && fast.next.next != null) {slow = slow.next;fast = fast.next.next;}// 步骤 2：反转链表的后半部分ListNode secondHalf = reverseList(slow.next);// 步骤 3：比较链表的前半部分和反转后的后半部分ListNode p1 = A;ListNode p2 = secondHalf;boolean result = true;while (result && p2 != null) {if (p1.val != p2.val) {result = false;}p1 = p1.next;p2 = p2.next;}// 步骤 4：恢复链表的原始结构slow.next = reverseList(secondHalf);return result;}// 反转链表的方法private ListNode reverseList(ListNode head) {ListNode prev = null;ListNode curr = head;while (curr != null) {ListNode nextTemp = curr.next;curr.next = prev;prev = curr;curr = nextTemp;}return prev;}public static void main(String[] args) {// 创建测试链表 1 -> 2 -> 2 -> 1ListNode head = new ListNode(1);head.next = new ListNode(2);head.next.next = new ListNode(2);head.next.next.next = new ListNode(1);PalindromeLinkedList solution = new PalindromeLinkedList();// 调用 isPalindrome 方法判断链表是否为回文结构boolean isPalindrome = solution.isPalindrome(head);System.out.println(isPalindrome);}
}

题目视图：

题目详解代码：

package Demo1_28;/*** Created with IntelliJ IDEA.* Description:* User：Lenovo* Date:2025-01-28* Time:20:08*/
// 定义链表节点类
class ListNode {int val;ListNode next;// 构造函数，用于初始化节点的值ListNode(int x) {val = x;next = null;}
}public class IntersectionOfTwoLinkedLists {// 查找两个链表相交的起始节点的方法public ListNode getIntersectionNode(ListNode headA, ListNode headB) {// 如果其中一个链表为空，直接返回 nullif (headA == null || headB == null) {return null;}// 初始化两个指针分别指向两个链表的头节点ListNode pA = headA;ListNode pB = headB;// 当两个指针不相等时，继续循环while (pA != pB) {// 如果 pA 到达链表 A 的末尾，将其指向链表 B 的头节点pA = pA == null ? headB : pA.next;// 如果 pB 到达链表 B 的末尾，将其指向链表 A 的头节点pB = pB == null ? headA : pB.next;}// 返回相交节点，如果不相交则返回 nullreturn pA;}public static void main(String[] args) {// 创建示例链表// 链表 A: 1 -> 2 -> 3 -> 6 -> 7ListNode headA = new ListNode(1);headA.next = new ListNode(2);headA.next.next = new ListNode(3);// 链表 B: 4 -> 5 -> 6 -> 7ListNode headB = new ListNode(4);headB.next = new ListNode(5);// 创建相交部分ListNode intersection = new ListNode(6);intersection.next = new ListNode(7);// 连接链表 A 和相交部分headA.next.next.next = intersection;// 连接链表 B 和相交部分headB.next.next = intersection;// 创建 IntersectionOfTwoLinkedLists 类的实例IntersectionOfTwoLinkedLists solution = new IntersectionOfTwoLinkedLists();// 调用 getIntersectionNode 方法查找相交节点ListNode result = solution.getIntersectionNode(headA, headB);if (result != null) {System.out.println("相交节点的值为: " + result.val);} else {System.out.println("两个链表不相交");}}
}

题目视图：

题目详解代码：

// 定义链表节点类
class ListNode {int val;ListNode next;// 构造函数，用于初始化节点的值ListNode(int x) {val = x;next = null;}
}public class LinkedListCycle {// 判断链表是否有环的方法public boolean hasCycle(ListNode head) {// 如果链表为空或者只有一个节点，肯定没有环if (head == null || head.next == null) {return false;}// 慢指针，初始指向头节点，每次移动一步ListNode slow = head;// 快指针，初始指向头节点的下一个节点，每次移动两步ListNode fast = head.next;// 当慢指针和快指针不相等时，继续循环while (slow != fast) {// 如果快指针到达链表末尾或者快指针的下一个节点是末尾，说明没有环if (fast == null || fast.next == null) {return false;}// 慢指针移动一步slow = slow.next;// 快指针移动两步fast = fast.next.next;}// 如果跳出循环，说明慢指针和快指针相遇了，链表有环return true;}public static void main(String[] args) {// 创建有环链表 1 -> 2 -> 3 -> 4 -> 2（环从节点 2 开始）ListNode node1 = new ListNode(1);ListNode node2 = new ListNode(2);ListNode node3 = new ListNode(3);ListNode node4 = new ListNode(4);// 构建链表连接关系node1.next = node2;node2.next = node3;node3.next = node4;// 形成环node4.next = node2;// 创建 LinkedListCycle 类的实例LinkedListCycle solution = new LinkedListCycle();// 调用 hasCycle 方法判断链表是否有环boolean result = solution.hasCycle(node1);System.out.println("链表是否有环: " + result);// 创建无环链表 1 -> 2 -> 3 -> 4ListNode nodeA = new ListNode(1);ListNode nodeB = new ListNode(2);ListNode nodeC = new ListNode(3);ListNode nodeD = new ListNode(4);// 构建无环链表连接关系nodeA.next = nodeB;nodeB.next = nodeC;nodeC.next = nodeD;// 再次调用 hasCycle 方法判断无环链表是否有环result = solution.hasCycle(nodeA);System.out.println("链表是否有环: " + result);}
}